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Riesz representation theorem

  1. Aug 1, 2008 #1
    I'm strugging with a portion of Rudin's proof.

    Quick statement of the bulk of the theorem:

    Let X be a locally compact Hausdorff space. Let A be a positive linear functional on Cc(X) (continous functions with compact support). Then (among other things), there exists a measure u() that represents A:

    A(f) = Integral(fdu) for every f in Cc(X).

    Now assuming that, he shows that u(K) for any compact set K is finite.

    He basically shows that u(K) <= A(f) for some f in Cc(X) with 0 <= f <= 1. This far I follow. But then he immediately concludes that therefore u(K) is finite for any compact K.

    Is there some basic property of positive linear functionals that makes them always finite? There is a somewhat similar proof in Rudin's Principles of Mathematical Analysis where he shows that the norm of linear functionals on finite dimensional vector spaces is finite. But that proof assumes a finite dimensional space. So I can't see how to apply that proof here.

    Any help is appreciated.
  2. jcsd
  3. Aug 2, 2008 #2
    I can use the compactness of K to choose an finite open cover {Vi} of K. Then we have:

    A(V1) + A(V2) = A(V1+V2), etc by linearity. Showing that this is >= u(K) is straightforward with the countable additivity of the measure. Since there are finite Vi, I get a finite sum very similar to Rudin's proof on the finite vector space in PMA.

    However, I'm still assuming that each A(Vi) is finite. But it seems like I could be on the right track. What's the justification that each A(Vi) is finite?

  4. Aug 2, 2008 #3


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    isn't this obvious? A(f) must be finite by the definition of a linear function, so u(K) <= A(f)< infinity.
  5. Aug 2, 2008 #4
    Let me make sure I understand.

    Since A: X -> R is a function, then for any f in X, A(f) must be finite by the definition of a function since A is well defined on its domain.

    Is that right?

    Also, Royden has a section about extended real-valued functions where apparently f(x)=infinity for some x is a valid definition. So I just need to assume that we aren't dealing with these extended functions here.

  6. Aug 2, 2008 #5


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    Yes, except that it is A:Cc(X)->R and f in Cc(X). The Riesz representation theorem is for real valued linear maps.
    Although, u(f) could be infinite if f>=0 doesn't have compact support but then it isn't in the domain of A anyway.
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